Question

1) A book consists of 264 pages. Parul had read 3/4 of the book till yesterday. How many more pages she has to read ?

2) Find the area of a square field of side 12 1/2m each.

3) If the cost of 7 1/2 kg apples is Rs. 600, find the cost of 1 kg apples.

4) Simplify : 5/7 x 1 2/3 + 1 5/7 x 1/3 = ?

Pls..... help

Answer 1

Answer:

1)total no of pages=264

pages read=264of3/4

264×3/4

66×3=198

pages left=264-198

66pages

2)side of sq. field=12×1/2

=6m

area=4×6=24m²

3)cost of7/2 kg of apples=600

cost of 1kg apples=7/2×600=2100

4)5/7×12/3+15/7×1/3

=60/21+5/7

=60+15/21

=75/21

=25/7

Answer 2

Step-by-step explanation:

1)

$no \: of \: pages \: is \: = 264$

$no \: of \: had \: been \: read \: are \: = \frac{3}{4} \: of \: 264 = \frac{3 \times 264}{4} = 3 \times 66 = 198 \: pgs$

$no \: of \: pgs \: has \: to \: read \: are \: 264 - 198 = 66pages$

2)

$area \: of \: square \: field \: is \: = s×s\: = (12 \frac{1}{2} ×12 \frac{1}{2} )) = 156.25$

3)

$cost \: of \: 7\frac{1}{2} kg \: apples \: is \: = 600rs$

$cost \: of \: a \: kg \: is \: = 600 \div 7\frac{1}{2} = 600 \div \frac{15}{2} = 600 \times \frac{2}{15} = 40 \times 2 = 80$

4)

$\frac{5}{7} \times 1\frac{2}{3} + 1 \frac{5}{7} \times \frac{1}{3} = \frac{5}{7} \times ( \frac{5}{3} + \frac{12}{7} ) \times \frac{1}{3} = \frac{5}{7} \times ( \frac{(35 + 36)}{21} ) \times \frac{1}{3} = \frac{5}{7} \times \frac{71}{21} \times \frac{1}{3} = \frac{355}{441}$